2024 Re-NEET Some Basic Concepts of ChemistryMolarity calculations Medium Question 11 of 49 in Some Basic Concepts of Chemistry

The amount of glucose required to prepare 250 mL of $\dfrac{M}{20}$ aqueous solution is: (Molar mass of glucose: 180 g mol$^{-1}$)

A 2.25 g
B 4.5 g
C 0.44 g
D 1.125 g

The answer is A.

Solution

Molarity $M=\dfrac{w_2\times1000}{M_2\times V(mL)}$. $\dfrac{1}{20}=\dfrac{w_2\times1000}{180\times250}\Rightarrow w_2=\dfrac{180\times250}{20\times1000}=2.25$ g.

Concept Molarity calculations

Some Basic Concepts of Chemistry in past papers

49 questions from this chapter have appeared across 23 exam years.